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STRUCTURE OF PHOSPHORUS ISOTOPES
By Prof. Lefteris Kaliambos (Natural Philosopher in New Energy) (August 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure of magic nuclei, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. Although phosphorus (P) has 23 isotopes from P-24 to P-46, only one of these isotopes is stable P-31; as such, it is considered a monoisotopic element. The longest-lived radioactive isotopes are P-33 with a half-life of 25.34 days and P-32 with a half-life of 14.263 days. All other have half-lives under 2.5 minutes, most under a second. The least stable is P-25 with a half-life shorter than 30 nanoseconds—the half-life of P-24 is unknown. WHY P-31 IS A STABLE NUCLIDE ' After a careful analysis of the structure of atomic nuclei I discovered that the beta decay is due to the fact that in unstable nuclei there exist single horizontal pn bonds of weak binding energy leading to the beta decay. For example in my paper STRUCTURE AND BINDING OF H3 AND He3 using the diagram of the structure of the H3 one sees that it is unstable because the two neutrons make single np bonds, while the He3 is stable because the one neutron between the two protons makes two np bonds per neutron. On the other hand the pp repulsions of long range lead to the instability when we have a small number of pn bonds per nucleon. For understanding the stability of P-31 with S = +1/2 you can read my STRUCTURE OF Si-31 AND P-31 . Also for comparing the structure of P-32 with S-32 you can read mySTRUCTURE OF P-32 AND S-32 . After a detailed study I found that the phosphorus isotopes are based on the structure of P-31 and P-30. For example the P-32 with S=+1 is based on the P-31 by adding the extra p17.with S=+1/2. Using the following diagrams of P-31 and p-30 we see that the P-31 is a stable nuclide because outside the core (Mg-24 with S=0 ) there exist the p13, n13, p14, n14, p15, n15 and n16 which make 5 strong vertical bonds and 7 weak horizontal bonds, while in the unstable P-30 there exist outside the parallelepiped 10 nucleons with 6 vertical bonds and 15 weak horizontal bonds which cannot overcome the pp repulsions of long range and lead to the decay. ' ' ' DIAGRAM OF STABLE P-31 WITH S =+1/2 ' ' ' ' . p12.........n12 '' -HP6 '' n''11........p11 ' ' . n10.......p10........n16''' ' +HP5 '' n15…...p9........n''9 ' '' . p''8.......... n''8 ….........p14 ' ' -HP4 p13.....n7........p7 ' ' . n6.........p6..........n14 ' '' +HP3 n13......p''5.......n''5 '' '' . '' '' p4......... n''4...........p''15' ' -HP2 n3.......p3 ' ' . n2.......p2''' ' +HP1 p1.......n1 ' NUCLEAR STRUCTURE OF P-33, P-35, P-37, P-39, P-41, P-43 AND P-45 WITH S =+1/2 After a careful analysis I found that the structure of the above unstable nuclides is based on the structure of P-31. Adding extra neutrons of opposite spins giving weak horizontal bonds in the structure of P-31with S=+1/2 we get the structure of the above unstable nuclides with the same S=+1/2. For example the P-45 has 14 more extra neutrons of opposite spins than those of P-31. Since the extra 14 neutrons give S=0 we get S=+1/2 +0 =+1/2 . ' ' NUCLEAR STRUCTURE OF P-29, P-27 AND P-25 WITH S =+1/2 Similarly in the absence of two neutrons of opposite spins in the structure of P-31 we get the structure of P-29 with S=+1/2 . Also in the case of p-27 we observe 4 absent neutrons of opposite spins , and in the case of P-25 we observe the absence of 6 neutrons of opposite spins. ' DIAGRAM OF P-30 WITH S=+1' ' . n10.......p10' ' +HP5 p9...........n9 ' ' . p8.........n8.......p15 ' ' -HP4 n7..........p7.......n15 ' ' . p12.........n6........p6.......n14' ' +HP3 n12.......p5.........n5.......p14 ' ' . n11........p4........n4........p13' ' -HP2 p11......n3..........p3.......n13 ' ' . n2........p2' ' +HP1 p1........n1 ' Here we see that the P-30 has 5 horizontal planes with 30 nucleons giving S =+1 because we observe 16 nucleons of positive spins at +Hp1, +HP3 and +Hp5, and 14 nucleons of negative spins at -HP2 and -HP4. That is S = 16(+1/2) +14(-1/2) = +1 ' ' STRUCTURE OF P-28 AD P-26 WITH S =+3 ' After a detailed analysis we found that the structure of the above unstable nuclides is based on the structure of P-30. Here the deuteron n13p13 changes the spin from S=-1 to S=+1 because it goes from the -HP2 to +HP5 for making horizontal bonds with the n9 and p10. Therefore under such a rearrangement of nucleons the new p-30 has S =+3. Then, in the absence of 2 neutrons of opposite spins we get the structure of P-28 with S=+3. In the same way the p-26 has the same spin of P-28 because the 2 more absent neutrons have opposite spins. ' ''' '''STRU CTURE OF P-24 with S = +1 In this case the absent 6 neutrons of P-24 with S=+1 have opposite spins with S=0. Since in the original structure of the P-30 we have S=+1 one gets S = +1 + 0 = +1. ' ' STRUCTURE OF p-34, P-36, P-38, P-40, P-42, p-44 AND P-46 Similarly the structure of the above unstable nuclides is based on the structure of P-30. For example the P-34 with S=+1 has 4 extra neutrons of opposite spins which make single horizontal bonds leading to the beta decay. However in the case of P-36 with S=-4 as in the case of P-28 the deuteron n13p13 after the new arrangement contributes to the change of spin from S =+1 to S=+3. In this case also all nucleons change the spin. For example we have -HP1, +HP2, -HP3 +HP4 and -HP5 giving S =-3. Then 2 extra neutrons of negative spins give the total S =-4, while 4 extra neutrons of opposite spins give the total number 36 of the P-36 with S=-4. Following the structure of P-36 with S=-4 we see that the 4 extra neutrons of positive spins give the spin S=-2 of the P-40. That is S =-4 +4(+1/2) =-2. Category:Fundamental physics concepts